Question

If the point p (2,2) is equidistant from the point a(-2,k) and b(-k,3), find k. also find the length of ap

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer:

The value of k is $\frac{15}{8}$ and the length of ap is $\frac{5\sqrt{41}}{8}$ units.

Step-by-step explanation:

It is given that point p (2,2) is equidistant from the point a(-2,k) and b(-k,3).

Distance formula:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$pa=pb$

$\sqrt{(-2-2)^2+(k-2)^2}=\sqrt{(-k-2)^2+(3-2)^2}$

$\sqrt{k^2 - 4 k + 20}=\sqrt{k^2 + 4 k + 5}$

Squaring both the sides.

$k^2 - 4 k + 20=k^2 + 4 k + 5$

$15=8k$

$k=\frac{15}{8}$

The value of k is $\frac{15}{8}$.

The length of ap is

$ap=\sqrt{(-2-2)^2+(k-2)^2}$

$ap=\sqrt{(-4)^2+(\frac{15}{8}-2)^2}$

$ap=\sqrt{16+(\frac{-1}{8})^2}$

$ap=\frac{5\sqrt{41}}{8}$

Therefore the length of ap is $\frac{5\sqrt{41}}{8}$ units.